$f(x)=\begin{cases} \text{sin}(x)&\text{for }0\leq x\leq\pi \\\\ \dfrac x\pi-1&\text{for }\pi<x\leq10 \end{cases}$ Find $\lim_{x\to \pi}f(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $\pi$ (Choice D) D The limit doesn't exist.
$x=\pi$ is on the boundary between the pieces of our piecewise function. In order to find $\lim_{x\to \pi}f(x)$, we need to find the one-sided limits. Let's find the limit as $x$ approaches $\pi$ from the left. We will use the fact that $f(x)=\text{sin}(x)$ for $x$ -values smaller than $\pi$. $\begin{aligned} &\phantom{=}\lim_{x\to \pi^-}f(x) \\\\ &=\lim_{x\to \pi^-}\text{sin}(x) \\\\ &=\text{sin}(\pi)&\gray{\text{Direct substitution}} \\\\ &=0 \end{aligned}$ Let's find the limit as $x$ approaches $\pi$ from the right. We will use the fact that $f(x)=\dfrac x\pi-1$ for $x$ -values greater than $\pi$. $\begin{aligned} &\phantom{=}\lim_{x\to \pi^+}f(x) \\\\ &=\lim_{x\to \pi^+}\dfrac x\pi-1 \\\\ &=\dfrac\pi\pi-1&\gray{\text{Direct substitution}} \\\\ &=0 \end{aligned}$ The one-sided limits are both equal to $0$. This means that $\lim_{x\to \pi}f(x)=0$.